<<

. 29
( 54 .)



>>

C to give,
1 ˆ
Ψ(C) = µ(trC ’ 3) (5.49)
2
Now using Equation (5.46) S is obtained with the help of Equations (5.19a)
and (5.20) as,
‚ Ψ(C)
+ pJC ’1
S=2
‚C
ˆ
‚trC
+ pJC ’1

‚C
‚ ’1/3
(IIIC C : I) + pJC ’1

‚C
’1/3 ’1/3’1
I ’ 1 IIIC IIIC C ’1 (C : I)] + pJC ’1
= µ[IIIC 3
’1/3
(I ’ 1 IC C ’1 ) + pJC ’1
= µIIIC (5.50)
3

The corresponding Cauchy stress tensor can now be obtained by using
Equation (4.45b) to give σ as,
σ = J ’1 F SF T
= µJ ’5/3 F (I ’ 1 IC C ’1 )F T + pF C ’1 F T
3
1
σ = µJ ’5/3 (b ’ 3 IbI)
= σ + pI ; (5.51)
where the fact that Ib = IC has been used again.
15
5.5 INCOMPRESSIBLE MATERIALS



We can now evaluate the Lagrangian elasticity tensor with the help
of Equations (5.10) or (5.11). The result can be split into deviatoric and
pressure components, C and C p respectively, as,

‚2Ψ ‚(JC ’1 )
‚S ‚S
C=2 = C + Cp; C=2 =4 ; C p = 2p
‚C ‚C ‚C‚C ‚C
(5.52)

With the help of Equations (5.22) and (5.31) these two components can be
evaluated for the neo-Hookean case de¬ned by Equation (5.49) after lengthy
but simple algebra as,
’1/3 1
’ 1 I — C ’1 ’ 3 C ’1 — I + 9 IC C ’1 — C ’1
1 1
C = 2µIIIC 3 IC III (5.53a)
3

C p = pJ[C ’1 — C ’1 ’ 2I ] (5.53b)

Note that the pressure component C p does not depend on the particular
material de¬nition being used.
The spatial elasticity tensor is obtained by the push forward type of
operation shown in Equation (5.14) as,

c = J ’1 φ— [C]; c p = J ’1 φ— [C p ]
ˆ ˆ
c = c +c p ; (5.54)

Performing this push forward operation in Equations (5.53a,b) gives,

c = 2µJ ’5/3 1 ICi ’ 3 b — I ’ 1 I — b + 1 IbI — I
1
(5.55a)
3 3 9

cp = p[I — I ’ 2i ] (5.55b)


EXAMPLE 5.5: Mooney“Rivlin materials
A general form for the strain energy function of incompressible rubbers
attributable to Mooney and Rivlin is expressed as,
µrs (IC ’ 3)r (IIC ’ 3)s

Ψ(C) =
r,s≥0

where IIC is the second invariant of C de¬ned as,
2
IIC = 1 (IC ’ IIC );

IIC = C : C
2

The most frequently used of this family of equations is obtained when
only µ01 and µ10 are di¬erent from zero. In this particular case we
have,
2
1
Ψ(C) = µ10 (IC ’ 3) + 2 µ01 (IC ’ IIC ’ 6)
16 HYPERELASTICITY




EXAMPLE 5.5 (cont.)
The equivalent homogeneous potential is obtained by replacing C by
ˆ
C in this equation to give,
ˆ ˆ ˆˆ
Ψ(C) = µ10 (trC ’ 3) + 2 µ01 [(trC)2 ’ C : C ’ 6]
1




5.5.3 NEARLY INCOMPRESSIBLE HYPERELASTIC
MATERIALS
As explained at the beginning of Section 5.5 near incompressibility is often
a device by which incompressibility can more readily be enforced within
the context of the ¬nite element formulation. This is facilitated by adding a
volumetric energy component U (J) to the distortional component Ψ already
de¬ned to give the total strain energy function Ψ(C) as,
Ψ(C) = Ψ(C) + U (J) (5.56)
where the simplest example of a volumetric function U (J) is,
U (J) = 2 κ(J ’ 1)2
1
(5.57)
It will be seen in Chapter 6 that when equilibrium is expressed in a varia-
tional framework, the use of Equation (5.57) with a large so-called penalty
number κ will approximately enforce incompressibility. Typically, values of
κ in the region of 103 ’104 µ are used for this purpose. Nevertheless, we must
emphasize that κ can represent a true material property, namely the bulk
modulus, for a compressible material that happens to have a hyperelastic
strain energy function in the form given by Equations (5.56) and (5.57).
The second Piola“Kirchho¬ tensor for a material de¬ned by (5.56) is
obtained in the standard manner with the help of Equation (5.22) and noting
that IIIC = J 2 to give,
‚Ψ
S=2
‚C
‚Ψ dU ‚J
=2 +2
‚C dJ ‚C
‚Ψ
+ pJC ’1
=2 (5.58)
‚C
where, by comparison with (5.46), we have identi¬ed the pressure as,
dU
p= (5.59)
dJ
17
5.5 INCOMPRESSIBLE MATERIALS



which for the case where U (J) is given by Equation (5.57) gives,
p = k(J ’ 1) (5.60)
This value of the pressure can be substituted into the general Equation (5.58)
or into the particular Equation (5.50) for the neo-Hookean case to yield the
complete second Piola“Kirchho¬ tensor. Alternatively, in the neo-Hookean
case, p can be substituted into Equation (5.51) to give the Cauchy stress
tensor.

EXAMPLE 5.6: Simple shear (ii)
Again we can study the case of simple shear for a nearly incompressible
neo-Hookean material. Using Equation (5.51) and the b tensor given
in Exercise 5.4 we obtain,
®2 2 
γ γ 0
3
 
1
’ 3 γ2
σ = µ° γ 0»
’ 1 γ2
0 0 3
where now the pressure is zero as J = 1 for this type of deformation.
Note that for this type of material there is no Kelvin e¬ect in the sense
that a volume-preserving motion leads to a purely deviatoric stress
tensor.


EXAMPLE 5.7: Pure dilatation (ii)
It is also useful to examine the consequences of a pure dilatation on a
nearly incompressible material. Recalling that this type of deformation
has an associated left Cauchy“Green tensor b = J 2/3 I whose trace is
Ib = 3J 2/3 , Equations (5.51) and (5.60) give,
σ = κ(J ’ 1)I
As expected a purely dilatational deformation leads to a hydrostatic
state of stresses. Note also that the isochoric potential Ψ plays no role
in the value of the pressure p.


Again, to complete the description of this type of material it is necessary
to derive the Lagrangian and spatial elasticity tensors. The Lagrangian
tensor can be split into three components given as,
‚(JC ’1 )
‚Ψ ‚p
+ 2JC ’1 —
C=4 + 2p (5.61)
‚C‚C ‚C ‚C
18 HYPERELASTICITY



The ¬rst two components in this expression are C and Cp as evaluated in
the previous section in Equations (5.53a,b). The ¬nal term, namely C κ , rep-
resents a volumetric tangent component and follows from U (J) and Equa-
tion (5.22) as,
‚p
C κ = 2JC ’1 —
‚C
d2 U ’1 ‚J
= 2J C—
dJ 2 ‚C
d2 U ’1
2
C — C ’1
=J (5.62)
2
dJ
which in the case U (J) = κ(J ’ 1)2 /2 becomes,
C κ = κJ 2 C ’1 — C ’1 (5.63)
Finally, the spatial elasticity tensor is obtained by standard push forward
operation to yield,
c = J ’1 φ— [C] = c + cp +c κ
ˆ (5.64)
ˆ
where the deviatoric and pressure components, c and cp respectively, are
identical to those derived in the previous section and the volumetric com-
ponent c κ is,
d2 U
’1
cκ = J φ— [C κ ] = J 2 I — I (5.65)
dJ
which for the particular function U (J) de¬ned in Equation (5.57) gives,
c κ = κJ I — I (5.66)

Remark 5.5.1. At the initial con¬guration, F = C = b = I, J = 1,
p = 0, and the above elasticity tensor becomes,
ˆ
c = c +c κ
= 2µ i ’ 1 I — I + κI — I
3
= κ ’ 2 µ I — I + 2µi (5.67)
3

which coincides with the standard spatially isotropic elasticity tensor (5.37)
with the relationship between » and κ given as,
2
» = κ ’ 3µ (5.68)
In fact all isotropic hyperelastic materials have initial elasticity tensors as
de¬ned by Equation (5.37).
19
5.6 ISOTROPIC ELASTICITY IN PRINCIPAL DIRECTIONS



5.6 ISOTROPIC ELASTICITY IN PRINCIPAL DIRECTIONS

5.6.1 MATERIAL DESCRIPTION
It is often the case that the constitutive equations of a material are presented
in terms of the stretches »1 , »2 , »3 in the principal directions N 1 , N 2 , and
N 3 as de¬ned in Section 3.5. In the case of hyperelasticity, this assumes that
the stored elastic energy function is obtainable in terms of »± rather than
the invariants of C. This is most likely to be the case in the experimental

<<

. 29
( 54 .)



>>